if $$ 2sin^{2}\theta+3sin\theta -2=0  $$ ,   $$  0^\circ<\theta<90^\circ $$ then value of  $$ \theta $$ is

$$2sin^{2}\theta+3sin\theta -2=0$$

$$=$$> $$2\sin^2\theta-\sin\theta\ +4\sin\theta-2=0$$

$$=$$> $$\sin\theta\ \left(2\sin\theta\ -1\right)+2\left(2\sin\theta-1\right)=0$$

$$=$$> $$\left(2\sin\theta\ -1\right)\left(\sin\theta\ +2\right)=0$$

$$=$$> $$2\sin\theta\ -1=0$$  (or)  $$\sin\theta\ +2=0$$

$$=$$> $$\sin\theta\ =\frac{1}{2}$$  (or)  $$\sin\theta\ =-2$$

$$\sin\theta\ =-2$$ is not possible

$$\therefore\ \sin\theta\ =\frac{1}{2}$$

$$=$$>  $$\theta\ =30^{\circ\ }$$ since  $$0^{\circ\ }<\theta<90^{\circ\ }\ $$

Hence, the correct answer is Option D